Given that (4x−x212)12
rth term of an expression is Tr+1=Crnxn−ryr Tr+1=Cr12(4x)12−r(−x212)r=Cr12(41)12−r(−12)r⋅x12−r−2r
⇒12−3r=0
⇒r=4
So term independent from x is T5=C412⋅481⋅34⋅44=C412⋅4434
We know that Crn=(n−r)!r!n!
=55×4436
They given that it is =k4436
So required value of k=55.